Example of CRE1 Miniproject Ver.2 Student Edition (1)

8/17/2019 Example of CRE1 Miniproject Ver.2 Student Edition (1)

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Example of BKF2453 Integrated Miniproject: 50,000 Metric Tonnes per

Annum Production of Cumene from Alkylation of Benzene

Instruction

The reaction stoichiometry undergoes as follows

1. C3H6 + C6H6 → C9H12

propylene (P) benzene (B) cumene (C)

2.

C3H6 + C9H12 → C12H18

P C p-diisopropyl benzene (PDIPB)

and the reaction kinetics are based on the power law model

= (i)

= (ii)

where both reaction rates are in kmol kgcat-1 s-1, C P is concentration of propylene in kmol m-

3, C B is concentration of benzene in kmol m-3 and C C is concentration of cumene in kmol m

-3.

The specific rate constants (m6 kgcat

-1 kmol

-1 s

-1) are given by

= � ∆ (iii)

= � ∆

(iv)

Subscript numbers denote the number of reaction and the values of the parameters can be

seen in Table 1

Table 1: parameters of the reaction between from 300 – 350oC and 30 – 40 bar.

Reaction Ai

(m6 kgcat-1 kmol-1 s-1)

∆E i (kJ kmol-1)

1 3.5 x 104 -1.04 x 105

2 2.9 x 106 -1.47 x 105

The reaction occurs in gas phase with the feed of pure benzene liquid at 28 oC and 1 bar and

the feed of propylene liquid at the same temperature and 12 bar (from the vapour pressure

graph, the gaseous propylene liquefied at 10.17 bar for the temperature of 28oC). However

propylene feed contains 4.8% propane (PA) which does not involve in any reaction with


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continuously as it will accumulate in the reactive unit. Thus, propylene will recycled

internally and propane will be incinerated so as to generate steam for the process.

The block flow diagram of the process proposed at level 2 can be seen in Figure 1. F is molar

rate of input, P is molar flow rate of output for the respective components where,

additionally, the subscript G denotes the outlet gas of unconverted propylene and propane

and the subscript PA denotes propane.

Figure 1

The prices of material involved in the components are as in Table 2.

Table 2: Physical property and prices of components

Components Purity (%) Boiling Points (oC) Price/ Cost

Propylene 95.2 -42.7 $167/kmol

Benzene 99.9 80.1 $267/kmol

Cumene 99.9 152.4 $1200/kmol

PDIPB Not necessary 203 $0.1/kW fuel

Propane Not necessary -46 $0.1/kW fuel

Process

FP, FPA

FB

PC

PPDIPB

FPA


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Table of Contents

1. Introduction ....................................................................................................................... 4

2. Level 2 Decision: Input-Output Structure of the Process Flowsheet ................................ 4

2.1. Mole Balances in Term of Extent of Reaction ........................................................... 5

2.1.1. Reaction Selectivity, Yield and Stoichiometry ................................................... 6

2.2. Economic Potential .................................................................................................... 9

3. Level 3 Decision: Recycle Structure of the Flowsheet ..................................................... 10

3.1. Mole Balances for the Reactor ................................................................................ 11

3.2. Reactor Design ......................................................................................................... 12

3.2.1. Thermal Effect: Adiabatic or not ..................................................................... 13

3.2.2. Reactor Design ................................................................................................. 14

3.2.3. Reactor Cost .................................................................................................... 16

3.3. Compressor Design and Costs ................................................................................. 16

3.3.1. Efficiency.......................................................................................................... 17

3.3.2. Annualized Installed Cost ................................................................................ 17

3.3.3. Operating Cost ................................................................................................. 17

3.4. Economical Potential ............................................................................................... 17

4. References ....................................................................................................................... 18

5. Appendix A

6. Appendix B

7. Appendix C

8. Appendix D


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1. Introduction

Conceptual design should normally start with decision on the mode of operation. Since the

BKF2453 students do not learn the process economics in attaining right production rate and

detail qualitative analysis in order to decide either batch or continuous, the mode of

operation for the projects has thus been set to continuous one. The process runs for 8000 hr

yearly with the remaining hours are for plant turn-around and cleaning.

2. Level 2 Decision: Input-Output Structure of the Process Flowsheet

Raw Material and Impurities Management

The available raw materials for the process must be sought out from local or overseas

market. Their prices are usually based on the place of origin (Biegler, Grossmann, &

Westberg, 1997; Poling, THompson, Friend, Rowley, & Wilding, 2008) (incurring shipment

cost) and purity. Nevertheless, in this project, they are all readily given for the sake of

brevity.

From the project instruction, the benzene liquid is fed to the process in pure form. Thus

there is no need to pre-treat anything. However, the propylene feed is having 4.8% propane.

Propane can remain in the feed to the reactor as the composition is relatively low and it is

considered inert to the reaction. However, propane will be separated from propylene and

taken out from the process.

Output Decision

The raw materials costs normally fall in the range from 33 to 85% of the total processing

cost, and these costs are essential to be estimated before any other detail is added to the

design (Douglas, 1988). At this level, the impurity comes from the fresh propylene feed

which is propane.

As the rule of thumb in a process design, it is desirable to recover more than 99% of all

valuable materials. Thus at this level of decision, the mole balances are calculated for the

overall system in which the limiting reactant is assumed to get fully converted due to the

recycle. The recycle streams will not appear because it comes out from and goes in again

into the process, except for the recycle and purge, which is not in this case.

As referred to the heuristics of the destination code in Table 5.1-3 of the Douglas textbook,

the destination of products and impurity of the process are classified based on the boiling

point at the atmospheric pressure of all species in the process as can be summarised in

Table 2 1


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Table 2.1 Destination Code for Toluene HDA Process

Components Boiling Point (oC) Destination Code

Propylene -42.7 RecyclePropane -46 Fuel

Cumene 152.4 Primary product

Benzene 111 Recycle

PDIPB 203 Fuel

Outlet components exit the process at the ambient condition. From the boiling point, the

purge stream of propane is in gas phase while both cumene and p-diisopropyl benzene exit

as in liquid phase. However, cumene is the main product and must exit separately at a

standard purity to a particular price which, in this case, it is 99.9%. Figure 2.1 shows the

streams of input and output materials of the process.

Figure 2.1 Input-Output Stream and the Respective Nomenclatures of the Benzene

Alkylation Process

2.1. Mole Balances in Term of Extent of Reaction

The benzene alkylation occurs in gas phase with a side reaction of p-diisopropyl benzene

Process

1 Propylene, propane

2 Benzene

3 Cumene

4 p-diisopropyl benzene

5 propane

Process

F FP , F FPA

F FB

F FPA

PC

PPDIPB

(a)

(b)


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Analysis of mole balance follows the extent of reaction method (Felder & Rousseau, 2005).

The symbols ξ 1 and ξ 2 are used to denote the extents of reaction for the first and second

reactions (Eq. (

2.1), respectively. The mole balance of all chemical species are generallycomputed using the correlation as follows

i = + ξ where i is molar flow rate of the species i and νI is the stoichiometric coefficient. Hence,the final flow rate from Figure 3.1 can be summarised as in Table 2.2.

Table 2.2 Mole Balance for the Second Level of Decision

Species Inlet Change Outlet

Benzene F FB - ξ 1 0 Propylene F FP - ξ 1 - ξ 2 0 Cumene 0 ξ 1 - ξ 2 PCp-diisopropyl benzene 0 ξ 2 PPDIPB

In total, other than PC , there are 5 unknown variables. Information available to solve thosevariables can be detailed based on the material and energy balance textbook as follows

(Felder & Rousseau, 2005)

Number of unknown variables in the process 4

2 Independent reactions +2

4 equation of extent of reactions -4

Composition of inert propane in the feed and purge (4.8%) -1

Total (Variables can be calculated by specifying any one of them) 1

From the degree of freedom, one variable, any of F FP, F FB, or PPDIPB , must first be specified

before the others can be calculated. In chemical reaction engineering, they can be expressed

in conversion of the limiting reactant, which is the reactor design variable and thus will be

the variable to calculate the mole balance.

2.1.1. Reaction Selectivity, Yield and Stoichiometry

The extent of reaction will be calculated in the stoichiometric manner employing theconversion and flow of limiting reactant. The limiting reactant can first be identified from the

scheme of reactor through selectivity analysis. Thus,

/ = =

=

= − 1 ( 2.2)


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where for cumene: C1 = 102.81, C2 = -8674.6, C3 = -11.922, C4 = 7.0048 x 10-6 and C5 = 2

thus, PP = 571,318 Pa or 5.71 bar. This means cumene exist in liquid phase at 30 bar, the

minimum pressure of the reaction. Benzene, however, by using the same method, its vapour

pressure at 300oC is 54.76 bar, reveals that its phase is gas. Hence, due to this pressure and

temperature conditions the concentration of benzene is always high at reactive gas phase to

have effective contact with propylene thus securing the maximum selectivity.

The selectivity equations (Equation 2.2) cannot simply quantify the difference between both

reactant feed flows required as taught in the BKF2453 subject. Hence, some detail analyses

by using the batch reactor algorithm is necessary to delineate the effects of feed ratios of

propylene:benzene. From calculations using Ordinary Differential Equation (ODE) in

Polymath (Appendix A) the results for various ratios can be summarised and plotted as

depicted in Figure 2.2. All selectivities clearly decrease as the conversion of propylene

increase. The graph also shows that propylene cannot be excess as it will render the

selectivity to be almost zero at nearly half of the limiting reactant mole and thus high

selectivity is attributed to excess benzene whereby the ratio of 1:4 shows the maximum. It

also implies that the limiting reactant of the process will be propylene.

Figure 2.2 selectivity as functions of propylene conversion at various molar feed ratio,

propylene:benzene

0.00E+00

1.00E+02

2.00E+02

3.00E+02

4.00E+02

5.00E+02

6.00E+02

7.00E+02

8.00E+02

9.00E+02

1.00E+03

0 0.2 0.4 0.6 0.8 1 1.2

S e l e c

t i v i t y

Xp

1:1

1:2

2:1

1:4


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The yield is the fraction of propylene converted in the reactor that corresponds to the

cumene flow at the reactor outlet. Also, this cumene is all recovered and taken (assuming

100% separation efficiency for the second level of decision) from the process. Using the

same algorithm for the selectivity (Appendix A), the resulting yield of the reaction can be

plotted as shown in Figure 2.3. The results of yield from various feed ratios show a good

agreement with the selectivity’s whereby the excess propylene clearly declines the yield to

zero before XP = 0.6. The feed with equal molar ratio exhibits significant drop up to 0.85 of

yield. The molar ratios of 1:2 and 1:4 do not have much difference, implies that further

excess of benzene is no longer sensitive to yield enhancement. Therefore the feed ratio,

propylene:benzene, of 1:2 is probably preferable for this process.

Figure 2.3 Reaction yields against the conversion of propylene for various feed ratio,

propylene:benzene

The equation of the yield against the conversion of propylene for the P:B feed ratio of 1:2 (as

exhibited in Figure 2.3) can be derived by using the Regression function in Polymath on thedata of the results from the batch reactor algorithm (Appendix B). The equation of power

with 0.99 R-square fits the correlation as follows

Y = 1 – 0.0193 X P1.6321 ( 2.5)

For a production of P mol/hr (52 08 kmol/hr) the propylene fed to the process F must be

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

1.05

1.1

0 0.2 0.4 0.6 0.8 1 1.2

Y i e l d

Xp

1:11:2

2:1

1:4


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∴ = .. ( 2.7)If a fraction Y of propylene is converted to cumene, a faction 1-Y must be lost to p-

diisopropyl benzene. From the stoichiometry of the reactions (Equation 2.1), the amount of

PDIPB , PPDIPB , must be

= 1 − = 1 − ( 2.8)The benzene input is identical to propylene input because benzene also involves indirectly

with the second reaction. Hence,

= ( 2.9)In summary, the above variables can be calculated for respective streams as tabulated in

Table 2.3.

Table 2.3 Stream Table of Mole Balance of Benzene Alkylation Process

Component 1 2 3 4 5

Propylene PC /Y 0 0 0 0

Propane 0.0480.952 0 0 0 0.048

0.952 Benzene 0 PC /Y 0 0 0

Cume 0 0 PC 0 0

P-diisopropyl benzene 0 0 0 1 − 0

Temperature (oC) 28 28 28 28 28

Pressure (bar) 12 1 1 1 1

Where Y = 1 – 0.0193 X P1.6321

;

2.2. Economic Potential

Since the practical values of the design variables depend on the process economics, the

stream costs are calculated where all of the costs of all raw materials and product streams

equated in term of design variables. The potential of economics (EP) at the second level is

therefore

EP ($/yr) = Cumene Value + Fuel Value of PDIPB + Fuel Value of Propane –

Propylene Cost – Benzene Cost. ( 2.10)

The prices of materials and fuel value for cost estimation are as tabulated in Table 3.4

Table 2.4 Price Data for Benzene Alkylation


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Power of fuel can be calculated from the enthalpies of combustion of PDIPB and propane: -

∆HC-PDIPBo = 6.82 kJ/kmol (Turton, Bailie, Whiting, & Shaeiwitz, 2008) and - ∆HC-PA

o = 2.2

kJ/kmol (Poling et al., 2008).

The economic potential in Eq 3.12 would finally be

= $1200 + $0.1 −∆ + −∆ .. −$267

−

$167 ( 2.11)where Y can be calculated from Equation 2.5.

Figure 2.4 shows the result of economic potential for the benzene alkylation process. All

range of conversion yields positive potential and the minimum is $902 mil per year at X P = 1.

The highest profit is indicated by the X P = 0.

As this economic potential is the annual profit that does not have to pay anything for capital

costs and utilities costs, all the analyses are just catering the sale of products (including

byproducts) and the purchasing of raw materials. The next level of decisions would use the

positive range obtained from this level.

Figure 2.4 Economic Potential – Level 2

3. Level 3 Decision: Recycle Structure of the Flowsheet

At the third level, the recycle stream is focused including the reactor. A compressor, as the

9.00E+08

9.02E+08

9.04E+08

9.06E+08

9.08E+08

9.10E+08

9.12E+08

9.14E+08

0 0.2 0.4 0.6 0.8 1 1.2

$ / y r

Xp


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Figure 3.1 Benzene Alkylation Process with Liquid and Gas Recycle Streams

3.1. Mole Balances for the Reactor

Before solving the mole balance, the unknown variables must be ensured that they can be

solved by using specifications and equations available from the process. The degrees of

freedom of 2 recycle junctions and a reactor based on Figure 3.1 can be tabulated as in

Table 3.1.

Table 3.1 Degree of freedom analyses for unit and junctions appeared at level 2 decision

Unit or junction 1 2 Reactor

Unknown variables F FP F PA , F PO and F P: 4 F FB F B and F B0: 3 F P0 , F PA , F P , F B0 , F B and

PPDIPB: 6

Independent reactions - - 2Balance F FP +F P = F P0: -1 F FB + F B = F B0: -1 Independent extent of

reactions: -4

Feed condition Purity percentage: -1 - Molar feed ratio, ΘB: -1Information from the

2nd level of decision

F FP = PC /Y: -1 F FB = PC /Y: -1 F PA: -1

Variables to be

specified

1 ( variable X P) 1 ( variable X P) 2 ( variable X P and F P0)

From the analysis, the solution must start at Junction 1 as it has the lowest degree and it

deals with the limiting reactant. Thus,

F FP +F P = F P0

SeparatorsReactor

Benzene recycle

Benzene

Feed

Propylene feed

Propylene

Recycle

Cumene

PDIPB

PC

PPDIPB

F P, PC , F PA

F B

F FB

F FP and F PA

F B0

F P0 , F PA

RG and y PH

1

2

Propane

F P

F B , PPDIPB

F PA

Compressor


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3.2 Block flow diagram of the benzene alkylation at the 2nd

level of decision: recycle

structure.

3.2. Reactor Design

Since both rate laws are the power law models and the reaction occurs irreversibly in

gaseous phase (assuming the mass transfer limitation is deemed negligible), the right and

economical reactor is thus the pack bed reactor as can be clearly seen from the plot of the

Levenspiel.

Based on the selectivity analysis, propylene feed must be limited at any cost. High selectivity

can be sustained if the concentration of cumene is low. This probably can be attained by

having continuous exit flow of cumene along the reactor length. In fact, it is critical as the

reactor has potential of distribution problem in the catalyst packing due to the multiplicity ofphases. The reactor scheme suggestion can be illustrated in 3.3 where the liquid phase will

flow downward of the catalyst packing.

P, B and PA

Separator

sReactor

Compressor

Benzene recycle

Benzene Feed

Propylene feed

Propylene

Recycle

Cumene

PDIPB

PC

PPDIPB

F P0 (1-X)

F P (Θ B -X)

F FB

F FP and F PA

Θ BF P

F P0

RG and y PH

1

2

Propane

F P0 (1-X)

F P (Θ B -X)

F PA


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3.2.1. Thermal Effect: Adiabatic or not

The thermal effect of the reactor can be observed from the energy balance calculation via

the plot of reaction temperature versus propylene conversion. From various references,

heat capacities of the involving components are not unique and changed with respect to

absolute temperature as tabulated in Table 3.2.

Table 3.2 Heat capacities of components:

Components C1 x 1e-5 C2 x 1e-5 C3 x 1e-3 C4 x 1e-5 C5

Benzene* 0.44767 2.3085 1.4792 1.6836 677.66

Propylene* 0.43852 1.506 1.3988 0.74754 616.46

Propane* 0.5192 1.9245 1.6265 1.168 723.6

Cumene* 1.081 3.7932 1.7505 3.0027 794.8

C1 C2 C3 C4 C5

PDIB** -30.99e-5 1.01728 -6.0 e-4

1.312e-7

* = + // + //

the unit is J/kmol – from Perry Handbook 8

th

ed.

**

= 1 + 2 + 3

+ 4, the unit is J/mol – from chemical property website

http://chemeo.com/cid/56-152-1

In using ODE of Polymath, only one dependant variable is allowed for one program. In

particular, for the batch reactor design algorithm, the variable is time. If differential

equation of enthalpy against absolute temperature is inputted, the program will then not

allow running. Normally, Matlab software is used to solve it. Since this BKF2453 project

encourages students to use Polymath and the range of reaction temperature is just 50oC, the

average heat capacities are used.

The previous algorithm is again employed here with the addition of heat effect, (Appendix

C), the differential equation, dT/dt (for the batch reactor – Equation T8-1.H page 477, Fogler

Textbook 4th

Ed with exclusion of the heat exchange term) using the average heat capacities.

The result of the propylene conversion against the reaction temperature can be seen in

Figure 3.4. The batch reactor algorithm is used here because the temperature is plotted

against conversion and not against the reactor size variable such as retention time or volume

or weight of catalyst. The temperature increase is only 0.3

o

C.


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Figure 3.4 The conversion versus reaction temperature for adiabatic reactor

The X-T plot in Figure 3.4 shows temperature increase due to the exothermic of the reaction

and the final temperature is not too high. There is no run away reaction occurs at anytemperature. Thus, adiabatic condition is preferable for the reactor of choice, the pack bed

reactor because of simpler design and cost incurred.

3.2.2. Reactor Design

The pack bed reactor is designed by using the desired production rate where the feed is

changed by the conversion of propylene following Equation 3.1. The algorithm of reactor for

multiple reactions is thus employed here with several molar flow rates calculated from theconversion specified in the range of 0 1.

The mole balance of a pack bed reactor is

= ′ where i = P, B, C and PDIPB

The net rate laws are calculated as following

= for Reaction 1 = for Reaction 2where


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=

∑ where -∆HRxn(T) = Y [-∆HRxn1(T)] + (1-Y)[ -∆HRxn2(T)] which Y is reaction

yield as Y = 1 – 0.0193 X P1.6321 and X P = (F P0 – F P)/F P0

is heat capacity for respective components of which all change with temperature.However, in this project average heat capacities between 300 – 350 oC are used in

Polymath Educational Edition.

Stoichiometry:-

From the selectivity analysis, propylene was identified as the limiting reactant

= where C T0 = P0/(RT ); F T = ∑ F i and y = P/P0Pressure drop:-

=

where F T0 = 3F P0 as for F B0 = 2F P0

and = whilst =

+1.75

with ρ0 = 19.1 kg m-3, gC = 1, DP =0.01 m, µ = 2.914 x 10

-5 Pa.s, φ = 0.5 and G =FP0*MWP/Ac. Ac is the cross sectional area for 1.5” schedule 40 pipe, which is

approximately 0.0013 m2.

From various conversions, the weight of catalyst obtained from the algorithm (can refer

Appendix D) is plotted as can be seen in Figure 3.5. Besides, the volume of the reactor canbe obtained from the weight of the catalyst by using its bulk density.


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Figure 3.5 the weight of catalyst and volume for PBR against the propylene conversion.

Two trends of weight and volume plotted at the same line but their values are referred at

left and right axes, respectively.

3.2.3. Reactor Cost

The plug flow reactor cost estimation follows the shell and tube method. From the graph of

purchased cost versus tube surface area ¼” schedule 40 pipes in Figure 15-13 (Peters &

Timmberhaus, 1991), the following correlation (readily included inflation for 2014) was

obtained.

Purchased Cost,

C A0 = 125933V + 651508 ( 3.2)

3.3. Compressor Design and Costs

A compressor is required as the process has a gas recycle stream. Compressors are so

expensive that spares are seldom provided for centrifugal units (although reciprocating

compressors may have spares because of a lower service factor). In practice, one standby

compressor would also be purchased to accommodate for any failure and breakdown.

The design equation for the theoretical horsepower (hp) for the centrifugal gas compressor

(for various pressure loads) is as follows

53.03 101out

in in

in

P hp P Q

P

γ

γ

− × = −

(

3.3)

where Pin = lbf /ft3, Qin = ft

3/min (calculate from gas real law equation for RG) and γ = (C p /C v –

1)/(C p /C v ). The exit temperature from a compressor stage is

out out

in in

T P

T P

γ

=

(

3.4)

where the temperatures and pressure must be in the Rankine absolute units. Values of γ that can be used for first estimates are given in Table

T bl 3 3 V l f f i f


17/31

The compressor will be designed pressures Pin = 1 bar to Pout = 15 bar which is identical to

the feed of fresh propylene. The pressure is enough to liquefy the propylene flow. The flow

of both propylene and benzene liquids will be dosed by pumps equipped with non-return

valves.

3.3.1. Efficiency

At this level, the compressor has 90% of efficiency to account for fluid friction in suction and

discharge valves, ports, friction or moving metal surfaces, fluid turbulence, etc. The driver is

also assumed to have 90% of efficiency to account for the conversion of the input energy to

shaft work.

3.3.2. Annualized Installed Cost

The brake horsepower, bph, is obtained by introducing the compressor efficiency into Eq.

4.9:

bph = hp/0.9 (

3.5)

Then, Guthrie’s correlation can be used to calculate the installed cost for various types of

compressors:

Installed Cost = ( )0.82&

517.5 2.11280

c

M S bhp F

+

( 3.6)

where F c is the correction factor (F c = 1 in this case) and M&S (Marshall and Swift inflation

index) those are obtainable from the Peters’ textbook (Peters & Timmberhaus, 1991).

3.3.3.

Operating Cost

By dividing the brake horsepower by the driver efficiency (80%) , the utility requirement can

be calculated. Then from the utility cost ($0.1/kW for 2014) and using 8000 hr/yr, the

operating cost can be obtained.

3.4. Economical Potential

The previous economic analysis for the input-output structure considered only the streamcosts, i.e., products plus by-products minus raw material costs. At the second level decision

where the reactor system complete with the recycle streams, the previous economic

potential will be minus to the cost of reactor and catalysts, and the cost of multistage

compressor elevating pressure from 1 bar to 15 bar of propylene recycle. The result of the

l l ti


18/31

( ) ( ) ( )0.823 2&

EP ($/yr) EP ($/yr) Catalyst cost installation cost 517.5 2.11280

Compressor Operating Cost

c

M S W V bhp F

= − − − +

−

as shown from the graph in Figure 3.6, indicates that the optimum conversion of propylene,

at the stage where no separators are considered yet, is 0.6.

Figure 3.6 The Profitability of the Process at Level 3 Decision versus Propylene

Conversion.

4. References

Biegler, L. T., Grossmann, I. E., & Westberg, A. W. (1997). Systematic Methods of Chemical

Process Design: Prentice-Hall.

Douglas, J. M. (1988). Conceptual Design of Chemical Processes. New York: McGraw Hill.

Felder, R. M., & Rousseau, R. W. (2005). Elementary Principles of Chemical Processes (3 ed.).

New York: John Wiley & Sons.

Peters, M. S., & Timmberhaus, K. D. (1991). Mass Transfer and Reactor Equipment - Designand Costs. In J. J. Carberry, J. R. Fair, W. P. Schowalter, M. Tirrell & J. Wei (Eds.),

Plant Design and Economics for Chemical Engineers (4th ed.). New York: McGraw

Hill.

Poling, B. E., THompson, G. H., Friend, D. G., Rowley, R. L., & Wilding, W. V. (2008). Physical

and Chemical Data. In D. W. Green & R. H. Perry (Eds.), Perry's Chemical Engineers'

$800,000,000

$820,000,000

$840,000,000

$860,000,000

$880,000,000

$900,000,000

$920,000,000

0 0.2 0.4 0.6 0.8 1 1.2

$ / y r

Xp


19/31

APPENDIX A

ALGORITHM USING BATCH REACTOR DESIGN TO ANALYSE SELECTIVITY OF REACTIONS


20/31

Appendix A.txtAPPENDIX A# mole balance

d(Cp)/d(t) = rpCp(0) = 209.911 # propylened(Cb)/d(t) = rb # benzeneCb(0) = 419.822d(Cc)/d(t) = rc # cumeneCc(0) = 0d(Cpdib)/d(t) = rpdib # p-diisopropyl benzeneCpdib(0) = 0

# rate lawr1prime = k1prime * Cp * Cb

r2prime = k2prime * Cp * Ccrp = -r1prime - r2primerb = -r1primerc = r1prime - r2primerpdib = r2primek1prime = 3.5e4 * exp(-1.04e5 / R / T)k2prime = 2.9e6 * exp(-1.47e5 / R / T)R = 8.314T = 300 + 273

# Selectivity, Yield and ConversionS = (r1prime - r2prime + 0.00001) / (r2prime + 0.00001)Y = (r1prime - r2prime + 0.00001) / (r1prime + r2prime + 0.00001)Xp = 1 - Cp * Nt * P0 / (Cp0 * Nt0 * P)Xb = theta - Cb * Nt * P0 / (Cp0 * Nt0 * P)Nt = (Cp + Cb + Cc + Cpdib) * VV = 1Nt0 = (Cp0 + Cb0) * VP0 = (Cp0 + Cb0) * R * TP = (Cp + Cb + Cc + Cpdib) * R * TCp0 = 209.911

Cb0 = 419.822theta = Cb0 / Cp0

t(0) = 0t(f) = 3600


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APPENDIX B

REGRESSION RESULT OF CORRELATION BETWEEN YIELD AND CONVERSION OF PROPYLENE

Page 1 of 2


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Model: Y = 1-A*Xp^B

Nonlinear regression settingsMax # iterations = 300

Precision

General

Source data points and calculated data points

POLYMATH ReportNonlinear Regression (L-M) 03-May-2014

Variable Initial guess Value 95% confidence

A 0.008 0.0192825 7.782E-05

B 1.74 1.63208 0.0341333

R^2 0.9978368

R^2adj 0.9977975

Rmsd 2.869E-05

Variance 4.861E-08

Sample size 57

Model vars 2

Indep vars 1

Iterations 6

Xp Y Y calc Delta Y1 0 1 1. 0

2 0.2563921 0.9970792 0.9979085 -0.0008293

3 0.3868124 0.9952462 0.995908 -0.0006618

4 0.4707557 0.9939056 0.9943618 -0.0004562

5 0.5109773 0.9932113 0.9935546 -0.0003433

6 0.5870242 0.9917937 0.9919166 -0.0001229

7 0.6226019 0.9910784 0.9911019 -2.347E-05

8 0.6561975 0.9903694 0.990305 6.444E-05

9 0.7142212 0.9890606 0.9888671 0.0001935

10 0.7387246 0.988473 0.988237 0.000236

Page 2 of 2


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22 0.9225748 0.9832496 0.9830939 0.0001557

23 0.9282395 0.9830617 0.9829242 0.0001375

24 0.9334649 0.9828867 0.982767 0.0001197

25 0.9427443 0.9825721 0.9824865 8.558E-05

26 0.9468626 0.9824308 0.9823615 6.932E-05

27 0.9506714 0.9822993 0.9822455 5.376E-05

28 0.9541956 0.9821769 0.982138 3.891E-05

29 0.9604798 0.9819569 0.9819456 1.13E-05

30 0.9632795 0.9818582 0.9818596 -1.431E-06

31 0.9658745 0.9817664 0.9817798 -1.341E-0532 0.9682808 0.981681 0.9817057 -2.466E-05

33 0.9725834 0.9815276 0.9815728 -4.52E-05

34 0.9745053 0.9814589 0.9815133 -5.444E-05

35 0.9762894 0.9813951 0.9814581 -6.297E-05

36 0.9779461 0.9813358 0.9814067 -7.089E-05

37 0.9809142 0.9812295 0.9813145 -8.5E-05

38 0.9822423 0.9811821 0.9812732 -9.109E-05

39 0.9834765 0.9811381 0.9812348 -9.667E-05

40 0.9846237 0.9810974 0.981199 -0.0001016

41 0.9866818 0.9810248 0.9811349 -0.0001101

42 0.9876038 0.9809926 0.9811061 -0.0001135

43 0.9884613 0.9809628 0.9810793 -0.0001165

44 0.9892589 0.9809355 0.9810544 -0.0001189

45 0.990691 0.9808873 0.9810096 -0.0001223

46 0.9913332 0.9808662 0.9809895 -0.0001233

47 0.9919307 0.980847 0.9809708 -0.0001238

48 0.9924868 0.9808295 0.9809534 -0.0001239

49 0.9934859 0.9807996 0.9809221 -0.0001225

50 0.9939341 0.980787 0.980908 -0.00012151 0.9943514 0.9807758 0.9808949 -0.0001191

52 0.9947397 0.980766 0.9808828 -0.0001168

53 0.9954379 0.9807506 0.9808609 -0.0001103

54 0 9957513 0 9807448 0 980851 -0 0001062


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APPENDIX C

ALGORITHM USING BATCH REACTOR DESIGN TO ANALYSE HEAT EFFECT OF REACTIONS

Polymath DEQ Program

No TitleFile: D:\My Tasks\Courseworks\Sem 2 2013-2014\BKF2453 Chem Rxn Eng 1\Assessments\Mini Project\Appendix C pol


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File: D:\My Tasks\Courseworks\Sem 2 2013-2014\BKF2453 Chem Rxn Eng 1\Assessments\Mini Project\Appendix C.pol

# APPENDIX C# mole balanced(Cp)/d(t) = rpCp(0) = 209.911 # propylened(Cb)/d(t) = rb # benzene

Cb(0) = 419.822d(Cc)/d(t) = rc # cumeneCc(0) = 0d(Cpdipb)/d(t) = rpdipb # p-diisopropyl benzeneCpdipb(0) = 0

# rate lawr1prime = k1prime * Cp * Cbr2prime = k2prime * Cp * Cc

rp = -r1prime - r2primerb = -r1primerc = r1prime - r2primerpdipb = r2primek1prime = 3.5e4 * exp(-1.04e5 / R / T0) * exp(-1.04e5 / R * (1 / T0 - 1 / T))k2prime = 2.9e6 * exp(-1.47e5 / R / T0) * exp(-1.47e5 / R * (1 / T0 - 1 / T))R = 8.314T0 = 300 + 273

# Energy balance

d(T)/d(t) = (-rp * V) * (-delHrxn) / (Np * Cpp + Nb * Cpb + Nc * Cpc + Npdipb * Cppdipb + Npa * Cppa) # Energy balanceT(0) = 573.15Np = Cp * VNb = Cb * VNc = Cc * VNpdipb = Cpdipb * VNpa = 0.048 * Np / 0.942delHrxn = -99000 - (Y * Cpc + (1 - Y) * Cppdipb - Cpb - Cpp) * (T - T0)Cpb = 118301.05Cpp = 75645.53Cppa = 103155.16Cpc = 237763.42Cppdipb = 421721.13

# Selectivity, Yield and ConversionS = (r1prime - r2prime + 0.00001) / (r2prime + 0.00001)Y = (r1prime - r2prime + 0.00001) / (r1prime + r2prime + 0.00001)Xp = 1 - Cp * Nt * P0 / (Cp0 * Nt0 * P)Xb = theta - Cb * Nt * P0 / (Cp0 * Nt0 * P)

Nt = (Cp + Cb + Cc + Cpdipb) * VV = 1Nt0 = (Cp0 + Cb0) * VP0 = (Cp0 + Cb0) * R * TP = (Cp + Cb + Cc + Cpdipb) * R * TCp0 = 209.911Cb0 = 419.822


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APPENDIX D

ALGORITHM USING PACKED BED REACTOR DESIGN TO ESTIMATE THE WEIGHT OF CATALYST AND

VOLUME OF THE REACTOR

ECONOMIC

POTENTIAL

CALCULATIONS

AT

LEVEL

3

DECISIONS

Polymath DEQ Program

No TitleFile: D:\My Tasks\Courseworks\Sem 2 2013-2014\BKF2453 Chem Rxn Eng 1\Assessments\Mini Project\Appendix D#2 - PBR design.pol


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y g j pp g p

APPENDIX D-1 Mole Balance of PBR

d(Fp)/d(W) = rp # Propylene balanceFp(0) = 1.446681718d(Fb)/d(W) = rb # benzene

Fb(0) = 2.893363436d(Fc)/d(W) = rc # cumeneFc(0) = 0d(Fpdipb)/d(W) = rpdib # p-diisopropyl benzeneFpdipb(0) = 0Ft0 = Fp0 * 3

= 1 - 0.0193 * Xp ^ 1.6321p = (Fp0 - Fp) / Fp0

Fp0 = 1.446681718

Rate Lawr1prime = k1prime * Cp * Cbr2prime = k2prime * Cp * Ccrp = -r1prime - r2primerb = -r1primerc = r1prime - r2primerpdib = r2primek1prime = 3.5e4 * exp(-1.04e5 / R / T0) * exp(1.04e5 / R * (1 / T0 - 1 / T))k2prime = 2.9e6 * exp(-1.47e5 / R / T0) * exp(1.47e5 / R * (1 / T0 - 1 / T))

R = 8.314 # kJ/kmol/K0 = 300 + 273

StoichiometryCp = Ct0 * Fp / Ft * y * T0 / TCb = Ct0 * Fb / Ft * y * T0 / TCc = Ct0 * Fc / Ft * y * T0 / TCpdipb = Ct0 * Fpdipb / Ft * y * T0 / TFt = Fp + Fb + Fc + FpdipbCt0 = P0 / (1000 * R * T0) # kmol/m^3P0 = 30e5 # Pa

Energy balancedelHrxn = -99000 - (Y * Cpc + (1 - Y) * Cppdipb - Cpb - Cpp) * (T - T0)Cpb = 118301.05 # J/kmol mean heat capacity between 300-350CCpp = 75645.53 # J/kmol mean heat capacity between 300-350CCppa = 103155.16 # J/kmol mean heat capacity between 300-350CCpc = 237763.42 # J/kmol mean heat capacity between 300-350CCppdipb = 421721.13 # J/kmol mean heat capacity between 300-350C

d(T)/d(W) = rp * delHrxn / (Fb * Cpb + Fp * Cpp + Fc * Cpc + Fpdipb * Cppdipb + 0.0001) # Energy Balance for PBR(0) = 573 # K

Pressure drop using air propertiesd(y)/d(W) = -alfa / (2 * y) * Ft / Ft0 * T / T0(0) = 1

alfa = 2 * beta / (Ac * rhoB * P0)


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Appendix D‐2

Xp Y EP2 Ffp Fp0 Fb0 G alfa W (kg) V (m3)

0.000000001 1 911917967 0.014467 14466667 28933333 264957265 5.88798E+14 1.85E+04 7.4

0.00085 1 911917870 0.014467 17.01961 34.03922 311.714488 815.7470646 1.83E+04 7.3

0.001000001 1 911917841 0.014467 14.46666 28.93331 264.9570649 589.4782606 1.60E+04 6.404

0.010000001 0.999989 911912549 0.014467 1.446682 2.893363 26.49600216 5.956185192 1.44E+04 5.764

0.020000001 0.999967 911901171 0.014467 0.723357 1.446714 13.24829393 1.506131881 1.51E+04 6.02

0.100000001 0.99955

911685603 0.014473 0.144732 0.289464 2.65076614 0.065743789 1.51E+04 6.02

0.200000001 0.998604 911197040 0.014487 0.072434 0.144869 1.326637805 0.018169804 1.57E+04 6.276

0.300000001 0.997295 910518832 0.014506 0.048353 0.096706 0.885586389 0.008853173 1.66E+04 6.628

0.400000001 0.995674 909676778 0.01453 0.036324 0.072648 0.665271076 0.005421388 1.76E+04 7.044

0.500000001 0.993773 908685939 0.014557 0.029115 0.058229 0.533234732 0.003754896 1.89E+04 7.564

0.600000001 0.991615 907556315 0.014589 0.024315 0.04863 0.445329311 0.002807558 2.06E+04 8.24

0.700000001 0.989217 906294986 0.014624 0.020892 0.041784 0.382636367 0.002211117 2.31E+04 9.244

0.800000001 0.986591 904907135 0.014663 0.018329 0.036658 0.335697893 0.001807719 2.68E+04 10.7

0.900000001 0.983749 903396616 0.014706 0.01634 0.032679 0.299260198 0.001520047 3.46E+04 13.828

0.950000001 0.98225 902596284 0.014728 0.015503 0.031007 0.28394238 0.001405764 4.20E+04 16.78949

0.970000001 0.981636 902267834 0.014737 0.015193 0.030386 0.278261821 0.001364383 4.80E+04 19.216

0.980000001 0.981326 902101835 0.014742 0.015043 0.030086 0.275509419 0.001344527 5.25E+04 20.988

0.990000001 0.981014 901934655 0.014747 0.014896 0.029791 0.272813233 0.0013252 6.11E+04 24.44

1.000000001 0.9807

901766298 0.014751 0.014751 0.029503 0.270171576 0.001306382 2.15E+05 85.92

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